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Question Number 22874 by selestian last updated on 23/Oct/17

	Answered by $@ty@m last updated on 23/Oct/17
	$Solutionof 9. S_1 =((n(n+1))/2) S_2 =((n(n+1)(n+2))/6) S_3 ={((n(n+1))/2)}^2 ∴((S_3 (1+8S_1 ))/S_2 ^2 )=((n^2 (n+1)^2 .{1+8×((n(n+1))/2)})/(4×((n^2 ×(n+1)^2 ×(2n+1)^2 )/(36)))) =(({1+4n(n+1)}×9)/((2n+1)^2 )) =((9(4n^2 +4n+1))/((2n+1)^2 )) =9$
	$S o l u t i o n o f 9 .$ $S_{1} = \frac{n (n + 1)}{2}$ $S_{2} = \frac{n (n + 1) (n + 2)}{6}$ $S_{3} = {\frac{n (n + 1)}{2}}^{2}$ $∴ \frac{S_{3} (1 + 8 S_{1})}{S_{2}^{2}} = \frac{n^{2} {(n + 1)}^{2} . {1 + 8 \times \frac{n (n + 1)}{2}}}{4 \times \frac{n^{2} \times {(n + 1)}^{2} \times {(2 n + 1)}^{2}}{36}}$ $= \frac{{1 + 4 n (n + 1)} \times 9}{{(2 n + 1)}^{2}}$ $= \frac{9 (4 n^{2} + 4 n + 1)}{{(2 n + 1)}^{2}}$ $= 9$
	Answered by $@ty@m last updated on 23/Oct/17
	$Solution of 10. a_1 , a_2 , a_3 ... a_n are in HP ⇒(1/a_1 ), (1/a_2 ),...(1/a_n ) are in AP ⇒a_2 =((2a_1 a_3 )/(a_1 +a_3 )) ⇒a_1 a_2 +a_2 a_3 =2a_1 a_3 −−−(1) Similarly a_3 a_4 +a_4 a_5 =2a_3 a_5 −−−(2) Now (1/a_3 )−(1/a_1 )=(1/a_5 )−(1/a_3 ) ⇒(2/a_3 )=((a_1 +a_5 )/(a_1 a_5 )) ⇒2a_1 a_5 =a_1 a_3 +a_3 a_5 ×2 ⇒4a_1 a_5 =2a_1 a_3 +2a_3 a_5 ⇒a_1 a_2 +a_2 a_3 +a_3 a_4 +a_4 a_5 =4a_1 a_5 −−(3) {using (1) & (2) Looking at the pattern of (1) and (3) we can easily conclude that a_1 a_2 +a_2 a_3 +a_3 a_4 +.....+a_(n−1) a_n =(n−1)a_1 a_n ⇒k=n−1 Ans.$
	$S o l u t i o n o f 10 .$ $a_{1}, a_{2}, a_{3} . . . a_{n} a r e i n H P$ $\Rightarrow \frac{1}{a_{1}}, \frac{1}{a_{2}}, . . . \frac{1}{a_{n}} a r e i n A P$ $\Rightarrow a_{2} = \frac{2 a_{1} a_{3}}{a_{1} + a_{3}}$ $\Rightarrow a_{1} a_{2} + a_{2} a_{3} = 2 a_{1} a_{3} - - - (1)$ $S i m i l a r l y$ $a_{3} a_{4} + a_{4} a_{5} = 2 a_{3} a_{5} - - - (2)$ $N o w \frac{1}{a_{3}} - \frac{1}{a_{1}} = \frac{1}{a_{5}} - \frac{1}{a_{3}}$ $\Rightarrow \frac{2}{a_{3}} = \frac{a_{1} + a_{5}}{a_{1} a_{5}}$ $\Rightarrow 2 a_{1} a_{5} = a_{1} a_{3} + a_{3} a_{5} \times 2$ $\Rightarrow 4 a_{1} a_{5} = 2 a_{1} a_{3} + 2 a_{3} a_{5}$ $\Rightarrow a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + a_{4} a_{5} = 4 a_{1} a_{5} - - (3)$ ${u s i n g (1) & (2)$ $L o o k i n g a t t h e p a t t e r n o f (1) a n d (3)$ $w e c a n e a s i l y c o n c l u d e t h a t$ $a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . . + a_{n - 1} a_{n} = (n - 1) a_{1} a_{n}$ $\Rightarrow k = n - 1 A n s .$
	Answered by math solver last updated on 23/Oct/17

	Commented by math solver last updated on 23/Oct/17

	$a l t e r n a t i v e s o l u t i o n o f q . 10 . . . .$ $i f y o u f i n d d i f f i c u l t y i n u n d e r s t a n d i n g$ $t h e p a t t e r n (a n s . g i v e n b y s a t y a m b r o) :)$
	Commented by $@ty@m last updated on 23/Oct/17

	$N i c e . .$ ${I t}^{'} s t h e s y s t e m a t i c s o l u t i o n .$
	Commented by math solver last updated on 23/Oct/17

	$t h a n k s!;)$
	Commented by selestian last updated on 24/Oct/17

	$t h A n k s s i r s y s t e m a t i c i s e a s y t o$ $u n d e r s t a n d$
	Answered by Rasheed.Sindhi last updated on 25/Oct/17

	$You can't use 'macro parameter character #' in math mode$ $E a s y s o l u t i o n!$ $\frac{S_{3} (1 + {8 S}_{1})}{S_{2}^{2}} is constant for all n \in N$ $n = 1 \Rightarrow S_{1} = S_{2} = S_{3} = 1$ $\frac{S_{3} (1 + {8 S}_{1})}{S_{2}^{2}} = \frac{1 (1 + 8 \times 1)}{1^{2}} = 9$ $You need to evaluate only for$ $one value of n ({It}^{'} s sufficient)$ $and simple case is n = 1 .$
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